Question Detail

PHY 101 PHY 101 Past Questions Objective Question
Take CBT

Question

The position of a projectile travelling in two-dimensional space is given by x(t) = (2t² + 4t + 3) m and y(t) = (3t² - 5t + 2) m. Calculate the magnitude of the projectile's displacement and the average velocity between time interval t = 2 s and t = 5 s.
Options
A 58.0 m, 29 m/s
B 72.2 m, 24.1 m/s
Correct Answer
C 58.0 m, 10.31 m/s
D 29.0 m, 19.31 m/s
Correct Answer

Option B is the correct answer.

Detailed Explanation

At t=2: x=2(4)+4(2)+3=8+8+3=19, y=3(4)-5(2)+2=12-10+2=4. At t=5: x=2(25)+4(5)+3=50+20+3=73, y=3(25)-5(5)+2=75-25+2=52. Displacement Δx=73-19=54, Δy=52-4=48. Magnitude = √(54²+48²)=√(2916+2304)=√5220=72.25 m. Average velocity = displacement/time = 72.25/3 = 24.08 m/s ≈ 24.1 m/s.

Hint

Displacement = final position - initial position, avg velocity = displacement/time.

Question Info

Course: PHY 101
Topic: PHY 101 Past Questions
Type: Objective
Difficulty: Not specified
Question 1008 of 1250
Previous Course Details Next